Chartrand Mathematical Proofs 3e Exercise 5.45

Question

I am self-studying this book, and I'm not sure if there is a typo in this question, or there is a gap in my understanding. The question is:

Let $R(x)$ be an open sentence over a domain S. Suppose that $\forall x\in S,R(x)$ is a false statement, and that the set T of counterexamples is a proper subset of S. Show that there exists a subset W of S such that $\forall x\in W,R(x)$ is true.

If $R(x)$ on domain S is false $\forall x\in S$, how can the set T of counterexamples be a proper subset of S? Should the question maybe read "$\forall x\in T,R(x)$ is a false statement"? If the question appears to be mathematically correct as given, could someone give me a counterexample please?

Answer 1

Ok, so I think I was reading the question as "$\forall x \in S$", "R(x) is a false statement".

I think the question should be understood as: '$\forall x \in S, R(x)$' is a false statement. This is equivalent to "$\exists x \in S, \neg R(x)$"

Answer 2

The set of counterexamples is the set of $x$ such that $\lnot R(x)$ holds. You assume that $\forall x R(x)$ fails, which is not the same as $\forall x\lnot R(x)$ holding.

The assumption about $T$ being a proper subset ensures that some elements do satisfy $R(x)$, and you need to find some of them.